ch03
225 pág.

ch03


DisciplinaTransferência de Calor2.600 materiais15.290 seguidores
Pré-visualização42 páginas
r 0 r r r
r dr dr r dr drfk
= \u2212 \u2264 \u2264 = \u2264 \u2264\uf8eb \uf8f6\uf8eb \uf8f6\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
\ufffd
Hence, integrating both equations twice,
2
f 1 1
f 2
f f f f
dT qr C qr C
T ln r C
dr 2k k r 4k k
= \u2212 + = \u2212 + +
\ufffd \ufffd (1,2)
c 3 3
c 4
c c
dT C C
T ln r C
dr k r k
= = + (3,4)
The corresponding boundary conditions are:
) ( ) ( )f f 1 c 1r 0dT dr 0 T r T r= = = (5,6)
( )[ ]
1 1 2
c c
f c c c 2
r r r r r r
dT dTdTfk k k h T r T
dr dr dr \u221e
= = =
\u2212 = \u2212 \u2212 = \u2212
\uf8f6 \uf8f6\uf8f6\uf8f7 \uf8f7 \uf8f7\uf8f8 \uf8f8 \uf8f8 (7,8)
Note that Eqs. (7) and (8) are obtained from surface energy balances at r1 and r2, respectively. Applying
Eq. (5) to Eq. (1), it follows that C1 = 0. Hence,
2
f 2
f
qr
T C
4k
= \u2212 +
\ufffd (9)
From Eq. (6), it follows that
2
3 11
2 4
f c
C ln rqr C C
4k k
\u2212 + = +
\ufffd (10)
Continued...
PROBLEM 3.88 (Cont.)
Also, from Eq. (7),
2
31 1
3
1
Cqr qr
or C
2 r 2
= \u2212 = \u2212
\ufffd \ufffd (11)
Finally, from Eq. (8), 3 3 2 4
2 c
C C
h ln r C T
r k \u221e
\u2212 = + \u2212
\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb
 or, substituting for C3 and solving for C4
2 2
1 1
4 2
2 c
qr qrC ln r T
2r h 2k \u221e
= + +
\ufffd \ufffd (12)
Substituting Eqs. (11) and (12) into (10), it follows that
2 2 2 2
1 1 1 1 1
2 2
f c 2 c
qr qr ln r qr qrC ln r T
4k 2k 2r h 2k \u221e
= \u2212 + + +
\ufffd \ufffd \ufffd \ufffd
2 2 2
1 1 2 1
2
f c 1 2
qr qr r qrC ln T
4k 2k r 2r h \u221e
= + +
\ufffd \ufffd \ufffd (13)
Substituting Eq. (13) into (9),
( ) 2 22 2 1 2 1f 1
f c 1 2
q qr r qr
T r r ln T
4k 2k r 2r h \u221e
= \u2212 + + +
\ufffd \ufffd \ufffd (14)<
Substituting Eqs. (11) and (12) into (4),
2 2
1 2 1
c
c 2
qr r qr
T ln T
2k r 2r h \u221e
= + +
\ufffd \ufffd
. (15)<
(b) Applying Eq. (14) at r = 0, the maximum fuel temperature for h = 2000 W/m2\u22c5K is
( ) ( ) ( )
2 28 3 8 3
f
2 10 W m 0.006 m 2 10 W m 0.006 m 0.009 m
T 0 ln
4 2 W m K 2 25 W m K 0.006 m
× × × ×
= +
× \u22c5 × \u22c5
 
( )
( )
28 3
2
2 10 W m 0.006 m
300 K
2 0.09 m 2000 W m K
×
+ +
× \u22c5
( ) ( )fT 0 900 58.4 200 300 K 1458K= + + + = . <
(c) Temperature distributions for the prescribed values of h are as follows:
0 0.001 0.002 0.004 0.005 0.006
Radius in fuel element, r(m)
300
500
700
900
1100
1300
1500
Te
m
pe
ra
tu
re
, T
f(K
)
h = 2000 W/m^2.K
h = 5000 W/m^2.K
h = 10000 W/m^2.K
 
0.006 0.007 0.008 0.009
Radius in cladding, r(m)
300
400
500
600
Te
m
pe
ra
tu
re
, T
c(K
)
h = 2000 W/m^2.K
h = 5000 W/m^2.K
h = 10000 W/m^2.K
Continued...
PROBLEM 3.88 (Cont.)
Clearly, the ability to control the maximum fuel temperature by increasing h is limited, and even for h \u2192
\u221e, Tf(0) exceeds 1000 K. The overall temperature drop, Tf(0) - T\u221e, is influenced principally by the low
thermal conductivity of the fuel material.
COMMENTS: For the prescribed conditions, Eq. (14) yields, Tf(0) - Tf(r1) = 21 fqr 4k\ufffd = (2×108
W/m3)(0.006 m)3/8 W/m\u22c5K = 900 K, in which case, with no cladding and h \u2192 \u221e, Tf(0) = 1200 K. To
reduce Tf(0) below 1000 K for the prescribed material, it is necessary to reduce q\ufffd .
PROBLEM 3.89
KNOWN: Dimensions and properties of tubular heater and external insulation. Internal and external
convection conditions. Maximum allowable tube temperature.
FIND: (a) Maximum allowable heater current for adiabatic outer surface, (3) Effect of internal
convection coefficient on heater temperature distribution, (c) Extent of heat loss at outer surface.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conditions, (2) Constant properties, (3) Uniform
heat generation, (4) Negligible radiation at outer surface, (5) Negligible contact resistance.
ANALYSIS: (a) From Eqs. 7 and 10, respectively, of Example 3.7, we know that
( )2 2 22s,2 s,1 2 2 1
1
q r q
T T r ln r r
2k r 4k
\u2212 = \u2212 \u2212
\ufffd \ufffd (1)
and
( )2 22 1
s,1 ,1
1 1
q r r
T T
2h r\u221e
\u2212
= +
\ufffd
(2)
Hence, eliminating Ts,1, we obtain
( ) ( )2 2 2 2 22 2s,2 ,1 1 2 1 2
1 1 1
qr r 1 k
T T ln 1 r r 1 r r
2k r 2 h r\u221e
\u2212 = \u2212 \u2212 + \u2212
\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb
\ufffd
Substituting the prescribed conditions (h1 = 100 W/m2\u22c5K),
( ) ( )4 3 3s,2 ,1T T 1.237 10 m K W q W m\u2212\u221e\u2212 = × \u22c5 \ufffd
Hence, with Tmax corresponding to Ts,2, the maximum allowable value of \ufffdq is
6 3
max 4
1400 400
q 8.084 10 W m
1.237 10\u2212
\u2212
= = ×
×
\ufffd
with
( )
2 22
e c e
2
c 2
2 1
I L A II Re
q
LA
r r
\u3c1 \u3c1
\u3c0
= = =
\u2200
\u2212
\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb
\ufffd
( ) ( ) 1/ 21/ 2 6 32 2 2 2 2max 2 1 6
e
q 8.084 10 W m
I r r 0.035 0.025 m 6406 A
0.7 10 m
\u3c0 \u3c0
\u3c1 \u2212
×
= \u2212 = \u2212 =
× \u2126\u22c5
\uf8eb \uf8f6\uf8eb \uf8f6 \uf8ec \uf8f7\uf8ec \uf8f7 \uf8ec \uf8f7\uf8ed \uf8f8 \uf8ed \uf8f8
\ufffd <
Continued \u2026..
PROBLEM 3.89 (Cont.)
(b) Using the one-dimensional, steady-state conduction model of IHT (hollow cylinder; convection at
inner surface and adiabatic outer surface), the following temperature distributions were obtained.
0.025 0.027 0.029 0.031 0.033 0.035
Radius, r(m)
300
500
700
900
1100
1300
1500
Te
m
pe
ra
tu
re
, T
(K
)
h = 100 W/m^2.K
h = 500 W/m^2.K
h = 1000 W/m^2.K
The results are consistent with key implications of Eqs. (1) and (2), namely that the value of h1 has no
effect on the temperature drop across the tube (Ts,2 - Ts,1 = 30 K, irrespective of h1), while Ts,1 decreases
with increasing h1. For h1 = 100, 500 and 1000 W/m2\u22c5K, respectively, the ratio of the temperature drop
between the inner surface and the air to the temperature drop across the tube, (Ts,1 - T\u221e,1)/(Ts,2 - Ts,1),
decreases from 970/30 = 32.3 to 194/30 = 6.5 and 97/30 = 3.2. Because the outer surface is insulated, the
heat rate to the airflow is fixed by the value of q\ufffd and, irrespective of h1,
( ) ( )2 21 2 1q r r r q 15, 240 W\u3c0\u2032 = \u2212 = \u2212\ufffd <
(c) Heat loss from the outer surface of the tube to the surroundings depends on the total thermal
resistance
( )3 2
tot
i 3 2
ln r r 1
R
2 Lk 2 r Lh\u3c0 \u3c0
= +
or, for a unit area on surface 2,
( ) ( )2 3 2 2tot,2 2 tot
i 3 2
r ln r r r
R 2 r L R
k r h
\u3c0\u2032\u2032 = = +
Again using the capabilities of IHT (hollow cylinder; convection at inner surface and heat transfer from
outer surface through tot,2R\u2032\u2032 ), the following temperature distributions were determined for the tube and
insulation.
0.025 0.027 0.029 0.031 0.033 0.035
Radius, r(m)
1000
1040
1080
1120
1160
1200
Tu
be
 te
m
pe
ra
tu
re
, T
(K
)
delta =0.025 m
delta = 0.050 m
 
0 0.2 0.4 0.6 0.8 1
Dimensionless radius, (r-r2)/(r3-r2)
500
600
700
800
900
1000
1100
1200
In
su
la
tio
n 
Te
m
pe
ra
tu
re
, T
(K
)
r3 = 0.060 m
r3 = 0.085 m
Continued...
PROBLEM 3.89 (Cont.)
Heat losses through the insulation, ( )2q r\u2032 , are 4250 and 3890 W/m for \u3b4 = 25 and 50 mm, respectively,
with corresponding values of ( )1q r\u2032 equal to -10,990 and -11,350 W/m. Comparing the tube temperature
distributions with those predicted for an adiabatic outer surface, it is evident that the losses reduce tube
wall temperatures predicted for the adiabatic surface and also shift the maximum temperature from r =
0.035 m to r \u2248 0.033 m. Although the tube outer and insulation inner surface temperatures, Ts,2 = T(r2),
increase with increasing insulation thickness, Fig. (c), the insulation outer surface temperature decreases.
COMMENTS: If the intent is to maximize heat transfer to the airflow, heat losses to the ambient should
be reduced by selecting an insulation material with a significantly smaller thermal conductivity.
PROBLEM 3.90
KNOWN: Electric current I is passed through a pipe of resistance eR\u2032 to melt ice under
steady-state conditions.
FIND: (a) Temperature distribution in the pipe wall, (b) Time to completely melt the ice.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional radial conduction, (3)
Constant properties, (4) Uniform heat generation in the pipe wall, (5) Outer surface